Write each expression in the form $ca^pb^q$ Write each expression in the form $ca^pb^q$
c) $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$
\begin{align*}
&= \frac{a\left(\frac{2}{b}\right)}{1}*\frac{\left(\frac{a}{3}\right)}{3}=\dfrac{a^2\left(\frac{2}{b}\right)}{3}=\frac{a^2}{1}*\frac{2}{b}*\frac{1}{3}=\frac{2a^2}{3b}*\frac{b}{1}=\frac{2a^2b}{3}=\frac{2}{3}a^2b^1 
\end{align*} 
e) $\dfrac{a^{-1}}{(b^{-1})\sqrt{a}}$
\begin{align*}
&= \frac{1}{(b^{-1})a\sqrt{a}}=\frac{1b}{1a^1a^{\frac{1}{2}}}=\frac{1b^1}{1a^{\frac{2}{3}}}=1a^{\frac{-2}{3}}b^1
\end{align*}
These are my steps. Any corrections help.
 A: On what grounds did you move $b$ to the top on (c)? It's incorrect. You cannot just multiply by $\frac{b}{1}$ because it pleases you to do so. And, after you suddenly create a factor of $\frac{b}{1}$ ex nihilo, it would have cancelled with the denominator. So both the penultimate and antepenultimate equality signs are incorrect. 
$$\begin{align*}
\frac{a(\frac{2}{b})}{\frac{3}{a}} &= \frac{2a}{b}\frac{a}{3}\\
&= \frac{2}{3}\frac{a^2}{b}\\ 
&= \frac{2}{3}a^2b^{-1}.
\end{align*}$$
(d) is almost correct, except that $1+\frac{1}{2}=\frac{3}{2}$, not $\frac{2}{3}$. So the exponent of $a$ is incorrect.
A: I'll do the first in two ways, $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$, and we'll see what we think. First, I'm going to separate all the 'numbers'.


*

*$\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}= \dfrac{2a\frac{1}{b}}{3\frac{1}{a}} = \frac{2}{3}\dfrac{a\frac{1}{b}}{\frac{1}{a}}$
Now I'll do the $a$ terms.
$\frac{2}{3}\dfrac{a \frac{1}{b}}{\frac{1}{a}} = \frac{2}{3} a \frac{1}{b} \cdot \frac{a}{1} = \frac{2}{3}a^2 \frac{1}{b}$.
Finally, we know that $\frac{1}{b} = b^{-1}$. So we have $\frac{2}{3} a^2 b^{-1}$.

*Let's do it a different way. We can get rid of the $\frac{3}{a}$ on the bottom, as dividing by a fraction is the same as multiplying by its reciprocal.
$\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}} = a\left(\frac{2}{b}\right) \cdot \frac{a}{3} = a \frac{2a}{3b} = 2a^2 \frac{1}{3b} = \frac{2}{3} a^2 b^{-1}$.
And if I had any doubt, I could check my answer by plugging in some numbers and making sure that both sides give me the same number. 
A: If you remember that dividing by a fraction is the same as multiplying by its inverse, we get at once:
$$c)\,\,\frac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}=\frac{2a}{b}\frac{a}{3}=\frac{2a^2}{3b}=\frac{2}{3}a^2b^{-1}$$
$$(e)\,\,\frac{a^{-1}}{b^{-1}a^{1/2}}=a^{-1-1/2}\,b=a^{-3/2}\,b$$
